John is planning a party. It is a very hot day with air temperature of 88 degree

Question

John is planning a party. It is a very hot day with air temperature of 88 degrees. He gets home from shopping for the party at 1 PM, at which time he puts a pack of beer in the refrigerator. The temperature of the refrigerator is 45 degrees and the initial temperature of the beer is 88, same as room temperature. He then proceeds with other party preparations, but at 1:50 PM, fatigued and hot from the cleaning, he decides to have a beer. He takes one from the fridge but is disappointed by the fact that it has only cooled to 75 degrees. Worried, he does a quick calculation to predict the temperature the beer will have if he continues to keep it in the refrigerator until his guests arrive at 3 PM.

The predicted temperature of the beer kept in the refrigerator until 3:00 PM is __________ degrees. (Must be correct to +/- 0.1)

Unhappy with the answer, he moves the beer to the freezer. By the time the beer is moved, it is 1:55. The temperature of the freezer is 27. What is now the predicted temperature of the beer when his guests arrive?

The predicted temperature of the beer moved to the freezer at 1:55 and kept there until 3:00 PM is _________ degrees. (Must be correct to +/- 0.1)

Assume that at all times the temperature of the beer is described by the differential equation

Explanation / Answer

75 = 45 + (88-45)e^(-k*50)

k = .0072 min-1

Predicted temperature = 45 + (75-45)e^(-0.0072*70) = 45 + 0.468*30 = 63.12

Predicted temperature in the freezer,

= 27 + (75-27)*e^(-0.0072*65) = 27 + 48*0.495 = 57.048

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